OCR A Level Chemistry A: Basic concepts of organic chemistry (H432)

This study guide teaches section 4.1.1, Basic concepts of organic chemistry, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. It opens Module 4, Core organic chemistry, and every lettered outcome, 4.1.1(a) to (i), is covered below. H432 is the full A level; according to the specification, Module 4 is assessed in Synthesis and analytical techniques (02) and in Unified chemistry (03). This section has no PAG.

Other H432 topics are on the OCR chemistry hub; keep a record on the printable checklist and try the free diagnostics. Afterwards, use the revision notes and the practice questions.

Outcomes in 4.1.1

Outcome	Skill	Notes from the guidance
4.1.1(a)	IUPAC naming	First ten alkanes and alkyl groups; HSW8, HSW11
4.1.1(b)	General, structural, displayed, skeletal formulae	M4.2; COOH and COO; ring symbols
4.1.1(c)	Eight terms, from homologous series to unsaturated	Define homologous series only
4.1.1(d)	Predict formulae from a general formula	None
4.1.1(e)	Structural isomers	M4.2
4.1.1(f)	Homolytic and heterolytic fission	None
4.1.1(g)	Radicals and dots	No dots for diradicals
4.1.1(h)	Curly arrow meaning	No half arrows; HSW1, HSW8
4.1.1(i)	Mechanisms with arrows and dipoles	HSW1, HSW2, HSW8

Bond polarity is in the bonding and structure guide; empirical and molecular formulae (2.1.3 b) are in the amount of substance guide.

4.1.1(a): Naming organic compounds

IUPAC (the International Union of Pure and Applied Chemistry) sets the naming rules. A systematic name fits exactly one structure, so the name alone lets anyone draw the compound (HSW8, HSW11).

Stems and alkyl groups

Carbons	Alkane	Alkyl group
1	methane, CH₄	methyl, CH₃–
2	ethane, C₂H₆	ethyl, C₂H₅–
3	propane, C₃H₈	propyl, C₃H₇–
4	butane, C₄H₁₀	butyl, C₄H₉–
5	pentane, C₅H₁₂	pentyl, C₅H₁₁–
6	hexane, C₆H₁₄	hexyl, C₆H₁₃–
7	heptane, C₇H₁₆	heptyl, C₇H₁₅–
8	octane, C₈H₁₈	octyl, C₈H₁₇–
9	nonane, C₉H₂₀	nonyl, C₉H₁₉–
10	decane, C₁₀H₂₂	decyl, C₁₀H₂₁–

Prefixes and suffixes you meet in H432

Group	How it appears in the name	Example
alkene, C=C	suffix -ene with locant	but-1-ene
alcohol, –OH	suffix -ol (prefix hydroxy- when another group takes the suffix)	propan-2-ol
haloalkane	prefix fluoro-, chloro-, bromo-, iodo-	1-iodopropane
aldehyde, –CHO	suffix -al (always carbon 1)	ethanal
ketone, C=O	suffix -one	butanone
carboxylic acid, –COOH	suffix -oic acid	ethanoic acid
ring	prefix cyclo-	cyclopentane

Esters, acyl chlorides, amines, amides and nitriles are named when they arrive in Module 6, using the same rules.

The method

1. Find the longest carbon chain that contains the main functional group. That gives the stem.
2. Number from whichever end puts the suffix group (or the C=C) on the lowest-numbered carbon.
3. If that is still a tie, number to give the substituents the lowest set of numbers.
4. Name substituents in alphabetical order, ignoring di-, tri- and tetra-.
5. Use commas between numbers and hyphens between numbers and words.

The specification’s own example is CH₃CH₂CH(CH₃)CH₂OH, which is 2-methylbutan-1-ol: four-carbon chain including the C–OH, numbered from the OH end, with a methyl on carbon 2.

Worked example 1: structure to name

CH₃CH(CH₃)CH(C₂H₅)CH₂CH₃

Longest chain: five carbons, pentane. From the left the locants are {2, 3}; from the right {3, 4}. Use {2, 3}, and put ethyl before methyl.

Name: 3-ethyl-2-methylpentane.

CH₂=CHCH(CH₃)CH₂CH₃

Five carbons including the C=C. Number from the left so the double bond starts at C1; the methyl is on C3.

Name: 3-methylpent-1-ene.

CH₃CHBrCH₂CH(OH)CH₃

Five carbons. The –OH takes the suffix, so number from the right: OH on C2, Br on C4.

Name: 4-bromopentan-2-ol.

Worked example 2: name to structure

2,3-dichloro-3-methylhexane. Six carbons in a row; Cl on C2 and C3; CH₃ on C3; fill each carbon to four bonds with H.

CH₃–CHCl–C(CH₃)Cl–CH₂–CH₂–CH₃

Check: numbering from the other end would give {4, 4, 5}, which is higher than {2, 3, 3}, so the name is right.

4.1.1(b): Types of formula

Definitions are not required, but you must read and draw each type and pick the one that suits the context (HSW8).

• General formula: the simplest algebraic formula for any member of a series. Alkanes are CₙH₂ₙ₊₂.
• Structural formula: the minimum detail that shows how the atoms are arranged, such as CH₃CH₂CH₂CH₃ or CH₃(CH₂)₂CH₃ for butane. Write a carboxyl group as COOH and an ester group as COO.
• Displayed formula: every atom and every bond shown.
• Skeletal formula: hydrogen atoms on carbon chains are removed. Each line end and each corner is a carbon. Functional groups are drawn in full.

Cyclohexane is drawn as a plain hexagon. Benzene is a hexagon containing a circle.

Worked example 3: four formulae for butanoic acid

• Molecular formula: C₄H₈O₂ (empirical formula C₂H₄O).
• Structural formula: CH₃CH₂CH₂COOH.
• Displayed formula:

    H   H   H   O
    |   |   |   ||
H – C – C – C – C
    |   |   |    \
    H   H   H     O – H

• Skeletal formula: a zigzag of three lines. The left end is the CH₃ carbon. The final corner carbon carries a double-bonded O and an OH.

          O
          ||
    /\    /\
   /  \  /  \
       \/    OH

In a displayed formula every bond must appear, including O–H. Writing “C–OH” is not displayed.

4.1.1(c) and (d): Series, groups and classification

Homologous series: a series of organic compounds with the same functional group, where each member differs from the next by CH₂. This is the only definition in 4.1.1 that the specification says you must give.

Other terms you must use correctly:

• Functional group: the group of atoms responsible for a compound’s characteristic reactions.
• Alkyl group: formula CₙH₂ₙ₊₁, such as CH₃– or C₃H₇–. The letter R may stand for an alkyl group, or for any other part of a molecule not involved in the reaction.
• Aliphatic: carbon and hydrogen joined in straight chains, branched chains or non-aromatic rings.
• Alicyclic: an aliphatic compound arranged in non-aromatic rings, with or without side chains.
• Aromatic: contains a benzene ring.
• Saturated: single carbon–carbon bonds only. Unsaturated: contains multiple carbon–carbon bonds, including C=C, C≡C and aromatic rings.

The specification uses saturated and unsaturated only for carbon–carbon multiple bonds. It keeps the wider term “degree of saturation” for any multiple bond and for rings.

Compound	Aliphatic, alicyclic or aromatic?	Saturated?
hexane	aliphatic	saturated
cyclobutane	alicyclic (and aliphatic)	saturated
cyclohexene	alicyclic (and aliphatic)	unsaturated
methylbenzene	aromatic	unsaturated
propene	aliphatic	unsaturated

Worked example 4: using a general formula

(i) An alkane has Mᵣ = 100.0. Find its formula. Use C = 12.0 and H = 1.0. For CₙH₂ₙ₊₂, Mᵣ = 12n + 2n + 2 = 14n + 2.

14n + 2 = 100.0
14n = 98
n = 7

The alkane is C₇H₁₆, heptane.

(ii) An alcohol CₙH₂ₙ₊₁OH has Mᵣ = 88.0. Mᵣ = 12n + (2n + 1) + 16.0 + 1.0 = 14n + 18.

14n + 18 = 88.0
n = 5

The alcohol is C₅H₁₁OH.

(iii) The alkane with 15 carbons has 2(15) + 2 = 32 hydrogens: C₁₅H₃₂.

4.1.1(e): Structural isomers

Structural isomers share a molecular formula but differ in structural formula. They can differ in the carbon chain, in where a group sits on the chain, or in which functional group is present. Cyclopropane and propene, both C₃H₆, are an example of the last kind.

To find every isomer:

1. Work out what the formula allows (a ring or C=C, a halogen, an O).
2. Place the groups in every distinct position on the longest chain.
3. Shorten the chain by one carbon and repeat.
4. Name each structure. Two with the same name are one compound.

Worked example 5: isomers of C₃H₆BrCl

The formula fits CₙH₂ₙ₊₂ with two hydrogens replaced, so the molecule is a saturated three-carbon chain carrying one Br and one Cl. Place them in every distinct way.

Structure	Name
CH₃CH₂CHBrCl	1-bromo-1-chloropropane
CH₃CHClCH₂Br	1-bromo-2-chloropropane
CH₃CHBrCH₂Cl	2-bromo-1-chloropropane
BrCH₂CH₂CH₂Cl	1-bromo-3-chloropropane
CH₃CBrClCH₃	2-bromo-2-chloropropane

There are five structural isomers. The second and third rows are different compounds: the halogens have swapped carbons. In each, numbering from the CH₂X end gives the lower set, {1, 2}. In the fourth row both ends give {1, 3}. That is a tie, so the substituent first in the alphabet takes the lower number: 1-bromo-3-chloropropane, not 3-bromo-1-chloropropane.

4.1.1(f) and (g): Bond fission and radicals

A covalent bond is a shared electron pair, and it can break in two ways.

Homolytic fission: each bonding atom receives one electron from the pair. Two radicals form.

Heterolytic fission: one bonding atom receives both electrons. A cation and an anion form, and the more electronegative atom usually takes the pair.

A radical is a species with an unpaired electron. Show it with a dot on the atom that carries the electron. Dots are needed in every case where there is a single unpaired electron, such as Cl• and CH₃•. They are not needed for diradicals, such as an oxygen atom.

Worked example 6: writing the two kinds of fission

Homolytic fission of chlorine in ultraviolet light:

Cl–Cl → Cl• + Cl•

Heterolytic fission of the C–Br bond in bromocyclopentane. Bromine is more electronegative, so it takes both electrons:

C₅H₉–Br → C₅H₉⁺ + Br⁻

Radical mechanisms are written as a sequence of equations like these. Half curly arrows are not required (4.1.2 f).

4.1.1(h) and (i): Curly arrows and mechanisms

A curly arrow represents a pair of electrons moving. It either shows heterolytic fission (a bond breaking, with both electrons going to one atom) or the formation of a covalent bond.

Rules for drawing one:

• The tail starts at a bond, a lone pair or a negative charge. Never start it from an atom or a positive charge.
• The head points to where the pair ends up: an atom, or the space between two atoms where a new bond forms.
• Include any relevant dipoles, δ+ and δ−, on polar bonds.

Worked example 7: three arrow types

Bond formation from a lone pair. Ammonia accepts a proton:

H₃N:   +   H⁺   →   NH₄⁺
arrow: from the lone pair on N to the H⁺ ion

Heterolytic fission of a polar bond. Draw the dipole Cδ+–Brδ−. The arrow starts on the middle of the C–Br bond and ends on the Br atom.

Two arrows in one step. Hydroxide ion and 1-iodopropane:

         δ+   δ−
HO:⁻   CH₃CH₂CH₂–I
  arrow 1: from the lone pair on O⁻ to the δ+ carbon
  arrow 2: from the C–I bond to the I atom
products: CH₃CH₂CH₂OH + I⁻

The new O–C bond forms as the C–I bond breaks, so carbon never exceeds eight outer electrons. The reaction itself is in 4.2.2; here the point is the arrows.

Common errors

• Numbering from the wrong end: “pentan-4-ol” is pentan-2-ol.
• Listing substituents in number order, not alphabetical order.
• Counting one isomer twice because it was drawn flipped.
• Calling benzene saturated.
• Leaving the dot off a radical such as CH₃•.
• Starting a curly arrow from an atom or a δ+ end.

Next steps

Next, read the revision notes for fast recall; the practice questions come with marked answers.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.1.1 Basic concepts of organic chemistry, learning outcomes (a) to (i). Published by OCR.